I'm looking for the circuit to automatically switch the load (< 0.5 A) between the main power source (USB) and back-up (battery) without a specialized IC and without the functionality of charging the battery.

I found two options:

  • with 2 diodes
  • diode from the main source and switchable P-MOSFET from the battery enter image description here

I wonder if a similar solution can be implemented with MOSFETs only and came up with the following circuit:

enter image description here

Where VDC is the external power source, +BATT is voltage from battery, and VCC is the output voltage (automatically switched).

To my understanding (I'm not an EE) this should work as follows:

  • if VDC is connected:
    • the top P-MOSFET connects the VDC to VCC since the lower N-MOSFET shortens the gate of top P-MOSFET to GND and opens it
    • other P-MOSFETs are closed.
    • VDC = VCC
  • if VDC is not connected:
    • the VDC is pulled to GND, it opens the left bottom P-MOSFET, which pulls the top P-MOSFET to the VCC and closes it to prevent the back-flow from VCC to VDD
    • P-MOSFET connected to battery is open and connects the battery to load
    • VCC = +BATT

My questionas are:

  • Is it going to work? In case if VDC is USB and +BATT is 3xAAA batteries? Or if +BATT is a LiIon battery?
  • Is this solution any better in terms of saving energy than option with 2 diodes and/ one diode and mosfet? Are the voltage drops lower in such schematic?
  • Are there any other problems in this schematics except the hustle to find P-mosfers which can be open from small batteries voltage?
  • This circuit is going to work only if VDC > (+BATT + body voltage drop on right bottom P-MOSFET)?
  • \$\begingroup\$ As I understand, in given circuits the P-FETs function as reverse polarity protection, and because of that the drain and source should be reversed (interchanged) intentionally according to this \$\endgroup\$ Commented Nov 30, 2020 at 12:14

1 Answer 1


In principle this circuit should work, however here are two suggestions:

  • Remove the \$100k\Omega\$ resistor between VCC and the leftmost PFET. This resistor is slowing down the gate charging of the upper PFET during the VDC to BATTERY switchover. Meaning that, for a couple of miliseconds the battery might discharge into VDC if the latter \$VDC< V_{BAT} - V_{BD}\$.

  • Add some capacitance at the node VCC in order to account for voltage dips during the switchover.


The following simulation introduces a buffering capacitor \$C_1\$ which speeds up the gate capacitance charging of the upper PFET. Essentially the capacitor \$C_1\$ is charged when the VCC is active, and once the latter is disconnected, the capacitor discharges into the gate capacitance of the upper PFET.


  • \$\begingroup\$ I did doubt if that resistor was necessary, and my concern is if without it there could be potentially a situation when VCC is shortened to GND. Is it possible? Regarding the capacitor: there will be a voltage regulator downstream with the incoming capacitor. I'll count for that. \$\endgroup\$ Commented Nov 28, 2020 at 15:29
  • 1
    \$\begingroup\$ If you also want to cover this malfunction, you could keep the resistor but also add some additional voltage buffering for charging the gate capacitance of the upper mosfet. I will update my answer with some small simulation. \$\endgroup\$
    – vtolentino
    Commented Nov 28, 2020 at 17:18
  • 1
    \$\begingroup\$ Thank you! Probably, I also need to test this circuit in hardware. \$\endgroup\$ Commented Nov 30, 2020 at 12:09

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